a)Ta thấy:

\(\left(2x+\frac{1}{3}\right)^2\ge0\)

\(\Rightarrow\left(2x+\frac{1}{3}\right)^2-\frac{5}{6}\ge0-\frac{5}{6}=-\frac{5}{6}\)

\(\Rightarrow A\ge-\frac{5}{6}\)

Dấu "=" <=>x=-1/6

Vậy MinA=-5/6<=>x=-1/6

b)Ta thấy:\(\hept{\begin{cases}\left|2x+3\right|\\\left|y-\frac{1}{2}\right|\end{cases}\ge}0\)

\(\Rightarrow\left|2x-3\right|+\left|y-\frac{1}{2}\right|\ge0\)

\(\Rightarrow\left|2x-3\right|+\left|y-\frac{1}{2}\right|+\frac{3}{4}\ge0+\frac{3}{4}=\frac{3}{4}\)

\(\Rightarrow B\ge\frac{3}{4}\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}\left|2x-3\right|=0\\\left|y-\frac{1}{2}\right|=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{3}{2}\\y=\frac{1}{2}\end{cases}}\)

Vậy...

\(P=\left(2x+\frac{1}{x}\right)^2+\left(2y+\frac{1}{y}\right)^2\)

Ta có: \(2x+\frac{1}{x}\ge2\sqrt{2x+\frac{1}{x}}=2\sqrt{2}\)

\(\Rightarrow\left(2x+\frac{1}{x}\right)^2\ge8\)

\(\Rightarrow\left(2y+\frac{1}{y}\right)^2\ge8\)

Dấu \("="\) xảy ra \(\Leftrightarrow x=y=\pm\frac{1}{2}\)

Vậy \(P_{min}=16\Leftrightarrow x=y=\pm\frac{1}{2}\)

NT:(2x+1)^4>=0.Dấu ''='' xảy ra khi x=-1/2

=>(2x+1)^4-1>=-1.Dấu"=" xẩy ra khi x=-1/2

Vậy Min của biểu thức trên là -1

a: \(\left(2x+1\right)^4-1\ge-1\)

Dấu '=' xảy ra khi x=-1/2

b: \(\left(x^2-16\right)^2+\left|y-3\right|-2\ge-2\)

Dấu '=' xảy ra khi \(\left(x,y\right)\in\left\{\left(4;3\right);\left(-4;3\right)\right\}\)

\(A=\left(2x+\frac{1}{3}\right)^4-1\) . Có: \(\left(2x+\frac{1}{3}\right)\ge0\)

\(\Rightarrow\left(2x+\frac{1}{3}\right)^4-1\ge-1\)

Dấu = xảy ra khi: \(2x+\frac{1}{3}=0\)

\(\Rightarrow2x=-\frac{1}{3}\)

\(\Rightarrow x=-\frac{1}{3}:2=-\frac{1}{6}\)

Vậy: \(Min_A=-1\) tại \(x=-\frac{1}{6}\)

ta có a=3-x(1-2x)-(x-1)(x+2)=3-x+2x^2 -x^2-x+2=x^2-2x+5=(x^2 -2x+1)+4=(x-1)²+4< hoặc =4 <=>gtnn của a là 4 khi x-1=0 =>x=1

(x-1)^2+2(x-3) tinh

\(B=4x^2+5y^2-4xy+3x-y\)

\(\Leftrightarrow\left(4x^2-4xy+3x\right)+5y^2-y\)

\(\Leftrightarrow\left[4x^2-4x\left(y-\dfrac{3}{4}\right)+\left(y-\dfrac{3}{4}\right)^2\right]+5y^2-y-y^2+\dfrac{3}{2}y-\dfrac{9}{16}\)\(\Leftrightarrow\left(2x-y+\dfrac{3}{4}\right)^2+\left(4y^2-\dfrac{1}{2}y+\dfrac{1}{64}\right)-\dfrac{37}{64}\)

\(\Leftrightarrow\left(2x-y+\dfrac{3}{4}\right)^2+\left(2y-\dfrac{1}{8}\right)^2-\dfrac{37}{64}\ge\dfrac{-37}{64}\)

Vậy Min B = \(\dfrac{-37}{64}\) khi \(\left[{}\begin{matrix}\left(2x-y+\dfrac{3}{4}\right)^2=0\\\left(2y-\dfrac{1}{8}\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-y+\dfrac{3}{4}=0\\2y-\dfrac{1}{8}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-y+\dfrac{3}{4}=0\\2y=\dfrac{1}{8}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-\dfrac{1}{16}+\dfrac{3}{4}=0\\y=\dfrac{1}{16}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-11}{32}\\y=\dfrac{1}{16}\end{matrix}\right.\)

\(C=9y^2+2x^2-6y-6xy+5x-1\)

\(=\left(9y^2+6y-6xy\right)+2x^2+5x-1\)

\(=\left[9y^2+6y\left(1-x\right)+\left(1-x\right)^2\right]+2x^2+5x-1-1+2x-x^2\)\(=\left(3y-x+1\right)^2+\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{17}{4}\)

\(=\left(3y-x+1\right)^2+\left(x+\dfrac{3}{2}\right)^2-\dfrac{17}{4}\)

Vậy Min C = \(\dfrac{-17}{4}\) khi \(\left[{}\begin{matrix}\left(3y-x+1\right)^2=0\\\left(x+\dfrac{3}{2}\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3y-x+1=0\\x+\dfrac{3}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3y-\left(\dfrac{-3}{2}\right)+1=0\\x=\dfrac{-3}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=\dfrac{-5}{6}\\x=\dfrac{-3}{2}\end{matrix}\right.\)